jmc

Let $t=x+\sqrt{x^2+b^2}.$ Then $t-x=\sqrt{x^2+b^2},$ so $$(t-x{)}^2=x^2+b^2.$$Expanding, we get $$t^2-2tx+x^2=x^2+b^2,$$so $$x=\frac{t^2-b^2}{2t}.$$Hence, $$\begin{array}{rl}2(a-x)(x+\sqrt{x^2+b^2})& =2\left(a-\frac{t^2-b^2}{2t}\right)t\\ & =2at-t^2+b^2\\ & =a^2+b^2-(t-a{)}^2\\ & \le a^2+b^2.\end{array}$$Equality occurs when $t=a$ or $x=\frac{a^2-b^2}{2a},$ so the maximum value is $\boxed{a^2+b^2}.$